Question

Pls Help *MTH282*
Suppose the position vector of X and Y are (1,2,4) and (2,3,5), find the position vector of a point Z that bisect XY in the ratio 2:3
A 7i+12j+22k
B. 7i-12j+22k
C.frac{1}{7} (7i+12j+22k)
D.frac{1}{17} (7i-12j+22k)

Answer 1

Given:

Position vector of X =  (1,2,4)  = i + 2j + 4k

Position vector of Y =   (2,3,5) = 2i + 3j + 5k

To Find:

The position vector of a point Z that bisect XY in the ratio 2:3.

Solution:

• The position vector of a point Z which bisects XY in the ration n:1 , is given by,
• Z = $\frac{X + nY}{1 + n}$

1. x coordinate of Z :

• x =$\frac{1 + \frac{2}{3} * 2}{1+\frac{2}{3}}$  = 7/5

    2. y coordinate of Z :

• y = $\frac{2 + \frac{2}{3}*3}{1+ \frac{2}{3}}$ = 12/5

     3. z coordinate of Z :

• z = $\frac{4 + \frac{2}{3}*5}{1+ \frac{2}{3}}$ = 22/5

Therefore the position vector of Z that bisects XY in the ratio 2:3 is

$\frac{1}{5}$( 7i + 12j + 22k)